A box contains a mixture of small copper spheres and small lead spheres that are the same size. The total volume of both metals
1 answer:
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Answer:
Ionization energy of the metal and electron affinity of the nonmetal
Explanation:
An ionic bond is formed when a metal transfers electrons to a nonmetal.
M· + A ⟶ M⁺ + ·A⁻
The two main factors affecting this process are the
- Ionization energy of the metal
- Electron affinity of the nonmetal
1. Ionization energy
The lower the ionization energy of the metal, the more likely it is able to donate an electron.
2. Electron affinity
The higher the electron affinity of the nonmetal, the more likely it is to accept an electron,
3. Periodic trends
(a) Ionization Energy
Ionization energy increases from bottom to top and from left to right in the Periodic Table.
Thus, the atoms with the lowest ionization energy are in the lower left corner of the Periodic Table.
(b) Electron affinity
Electron affinity increases from bottom to top and from left to right in the Periodic Table.
Thus, the atoms with the highest electron affinity are in the upper right corner of the Periodic Table.
Answer:
431.38 mg protein / mL
Explanation:
This is an example of the <em>Kjeldahl method</em>, for nitrogen determination. All nitrogen atoms in the protein were converted to NH₃ which then reacted with a <u>known excess of HCl</u>. This excess was later quantified via titration with NaOH.
First we calculate the <u>total amount of H⁺ moles from HCl</u>:
- 0.0388 M HCl * 10.00 mL = 0.388 mmol H⁺
Now we calculate the <u>excess moles of H⁺</u> (the moles that didn't react with NH₃ from the protein), from the <u>titration with NaOH</u>:
- HCl + NaOH → H₂O + Na⁺ + Cl⁻
- 0.0196 M * 3.83 mL = 0.075068 mmol OH⁻ = 0.0751 mmol H⁺
Now we substract the moles of H⁺ that reacted with NaOH, from the total number of moles, and the result is the <u>moles of H⁺ that reacted with NH₃ from the protein</u>:
- HCl + NH₃ → NH₄⁺ + Cl⁻
- 0.388 mmol H⁺ - 0.0751 mmol H⁺ = 0.313 mmol H⁺ = 0.313 mmol NH₃
With the moles of NH₃ we know the moles of N, then we can <u>calculate the mass of N</u> present in the aliquot:
- 0.313 mmol NH₃ = 0.313 mmol N
- 0.313 mmol N * 14 mg/mmol = 4.382 mg N
From the exercise we're given the concentration of N in the protein, so now we <u>calculate the mass of protein</u>:
- 4.382 mg * 100/15.7 = 27.91 mg protein
Finally we <u>calculate the protein concentration in mg/m</u>L, <em>assuming your question is in 647 μL</em>, we first convert that value into mL:
- 647 μL *
0.647 mL
- 27.91 mg / 0.647 mL = 431.38 mg/mL